Let $x,y$ be vectors in $\mathbb{R}^n$ and let's use the notation $\hat x$ for the vector $x$ with its components sorted in increasing order.
The Hardy-Littlewood-Polya inequality states that
$$ x\cdot y \leq \hat x\cdot \hat y.$$
Let us also use the notation $xy\in\mathbb{R}^n$ to denote the coordinate-wise product of $x$ and $y$.
I conjecture that
$$
\frac{ ||xy||_p ||xy||_r}{||xy||_q}
\le
\frac{ ||\hat x\hat y||_p ||\hat x\hat y||_r}{||\hat x\hat y||_q}
$$
for all $1\le p\le q\le r$.
For $q=p$ and $q=r$, my conjectured inequality is true by the HLP inequality. Any ideas for a proof?

UPDATE: thank you for the quick answers. The counterexamples indeed work when negative coordinates for x and y are allowed. However, when all the coordinates of x and y are required to be positive, the conjecture seems to hold.

UPDATE 2: so the conjecture is totally false; see below for counterexamples.

The inequality is equivalent to the claim that $f(t)=\frac{\|xy\|_t}{\|\hat x\hat y\|_t}$ satisfies $f(s)f(t)\le f(1)$ for $s\le 1\le t$ ($1\le p$ is not really a restriction due to the possibility to raise to positive powers inside and outside, so only the ratios $p/q$ and $r/q$ really matter). Also sums can be replaced by averages. Now, as $s\to 0$, we have the geometric means in the limit, which do not feel the rearrangements, so $f(0+)=1$. Also, $f(\infty)=1$ if only the maxima match in the original arrangements. But $f(1)<1$ unless the orderings are exactly the same.